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  • $\begingroup$ I am not sure I understand why you think that the integral is zero based on the derivative? You are saying that eq. $(4)$ in the OP is not correct? Looking at my project as.a whole, it seems to be a consistent value. $\endgroup$ Commented Jun 13, 2020 at 22:09
  • $\begingroup$ I see, that's a great idea! Thanks! Is it actually clear that we can commute the limit and the integrals? I mean, I do it in my original post but your answer got me thinking about it. $\endgroup$ Commented Jun 14, 2020 at 10:00
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    $\begingroup$ I now actually think that we can't, but I might be wrong. Here is a simpler example. Consider integrand=x2^2*I12*I13*I24 integrated over $\tau_3$ and $\tau_4$ (MA can handle that easily). Applying your method, we can exchange the derivative and the integral without problem, but interchanging the limit and the integral gives an erroneous $0$. You can try it out with this code: D[integrand, x2]; Assuming[x1 > 0,Integrate[%, {\[Tau]3, -\[Infinity], \[Infinity]}, {\[Tau]4, -\[Infinity], \[Infinity]}]]; Limit[%, x2 -> 0, Direction -> "FromAbove"] $\endgroup$ Commented Jun 14, 2020 at 10:30
  • $\begingroup$ Note that the problem arises only at $0$. If you put $x_2 \to 10^{-6}$, then the limits can be interchanged so the script of the original question is still valid I guess. $\endgroup$ Commented Jun 14, 2020 at 10:39
  • $\begingroup$ Yes of course, I will also include the definition of $I_{12}$. $\endgroup$ Commented Jun 14, 2020 at 10:46